Homework 9 Solutions / Due: Fri19 November 2010
PAVEMENT DESIGN
CE361 HW9- 1 -
Please complete the exercises below completely and clearly. You may submit your assignment as a member of a group of CE361 students not to exceed three in size. Signatures of all group members must appear on the top page of the work submitted.
Dear Consultant:
Less than five years ago, an ethanol plant was built on a county road (CR600E) just off of SR361 east of Mythaca. The old gravel county road was replaced with a 2-lane 3-layer flexible pavement, designed to withstand the truck traffic associated with ethanol production. However, the pavement on CR600E is already starting to break up. The pavement designer blames the contractor for bad construction procedures; the contractor blames the designer for an inadequate design. You are hired to carry out the flexible pavement design “from scratch”, for comparison with the design that was actually used.
At your request, your staff has compiled the following preliminary calculations, assumptions, and design constraints.
Ethanol plant capacity: 90 million gallons per year. With a 5-day work week over 52 weeks per year,
= 346,154 gal/day
An acre of corn can be converted into 375 gallons of ethanol. Average corn yield is 50 bushels per acre. A bushel of corn weighs 56 lbs.
= 6.67 lb corn per gallon ethanol
If a 3-S2 semi-trailer is used to deliver the corn, will the corn load “cube out” (reach the truck’s payload volume limit) or “weigh out” (reach the truck’s payload weight limit)?
The load on the steering axle of the tractor of a 3-S2 corn truck– loaded or unloaded -- is 12K. The load from the tractor on its tandem drive axle is 3K. The weight of an empty semi-trailer on each of the two tandem axles in FTE Figure 9.7 is 2.5K. The payload in the semi-trailer will be evenly distributed over the two tandem axles in FTE Figure 9.7. An unloaded 3-S2 will weigh 12K+(3K+2.5K)+2.5K=20K. Ethanol taken from the plant will be carried in 3-S2 semi-trailer ethanol tankers with the same weight distributions on the axles as the 3-S2 corn trucks.
A semi-trailer is 98 in wide, 108 in high, and 53 ft long = 3895 cu ft.
3895 cu ft * = 3116 bu; 3116 bu * 56 lb/bu = 174,518 lbs. This far exceeds semi-trailer weight limit of 80,000 lbs total – (12,000+3000+2500+2500 lb) unloaded truck = 60,000 lb payload, so the weight of the corn is the constraint, not its volume.
Number of 3-S2 corn trucks serving the ethanol plant on an average day = = 38.48 39 3-S2 corn trucks per day, arriving loaded, and leaving unloaded at 20,000 lbs.
Number of 3-S2 ethanol tanker trucks serving the ethanol plant on an average day = = 38.46 39 tanker trucks per day, arriving empty at 20,000 lbs, and leaving loaded.
Check weight of 9000-gal load: 9000 gal * 6.6 lbs/gal ethanol = 59,400 lbs. This is less than 60,000-lb allowable payload. To simplify the ESAL calculations without compromising the design, assume 60K payload for both corn trucks and ethanol tankers.
Again, the payload in the semi-trailer will be evenly distributed over the two tandem axles in FTE Figure 9.7.
- ESAL calculations. Complete the table below for the design lane.
kips/axle / Axle
Type / Design lane
freq/da N(i) / ESAL/axle / Growth
rate r / Compounded
G / ESAL(i)/life / 3-S2
12.00 / S / 78 / 0.19753 / 0.034 / 38.44 / 153,971 / steering axle always
5.50 / T / 39 / 0.00075 / 0.034 / 38.44 / 294 / drive axle empty
2.50 / T / 39 / 0.00003 / 0.034 / 38.44 / 13 / trailer axle empty
35.50 / T / 39 / 1.30726 / 0.034 / 38.44 / 509,490 / drive axle full (corn)
35.50 / T / 0 / 1.30726 / 0.034 / 38.44 / - / drive axle full (tanker)
32.50 / T / 39 / 0.91829 / 0.034 / 38.44 / 357,895 / trailer axle full (corn)
32.50 / T / 0 / 0.91829 / 0.034 / 38.44 / - / trailer axle full (tanker)
Total Design Life ESALs = / 1,021,662
A.(10 points) Which lane – inbound to the plant or outbound – is the design lane, i.e., the lane with more ESALs? Explain how you determined the design lane and the load for each axle, for both empty and loaded trucks, on the design lane.
39 corn trucks each day enter full and leave empty. 39 ethanol tankers each day enter empty and leave full. Because they both weigh 20K empty and 60K full, the inbound and outbound roadway lanes will receive the same loadings. Either the inbound or the outbound lane can be the design lane. Choose one (and only one). Enter zero in the frequency column for the other direction. In the table above and the list below, the inbound lane is used as the design lane.
- Steering axle always = 12K. (78 per day on inbound lane)
- Drive axle empty = 3K+2.5K=5.5K. (39 per day on inbound lane)
- Trailer tandem empty = 2.5K. (39 per day on inbound lane)
- Drive axle (full corn trailerinbound) =3K+2.5K+(0.5*60K)=35.5K. (39 per day on inbound lane)
- Trailer tandem axle (full of corninbound)=2.5K+(0.5*60K)=32.5K. (39 per day on inbound lane)
B.(10 points) With one sample calculation for the single axle and one sample calculation for a tandem axle, show how you calculatedtheESAL/axle values.
(9.1) for single axle: ESAL(single)==0.1975.
(9.2) for tandem axle on flexible pavement: ESAL(tandem)=. If W=5500, ESAL=0.00075; if W=2500, ESAL=0.00003; if W=35,500, ESAL=1.307; if W=32,500, ESAL=0.9183.
C.(10 points) Show how you calculatedand applied the growth factor needed to determine ESALs over the lifetime of the pavement. Use a 25-year design life and a 3.4 percent per year growth rate in ethanol production.
(9.4) Lifetime ESAL = ESAL/yr * ==38.436.
D.(10 points) How many ESALS should the pavement be designed for?
Sample calc for Steering axle empty: ESAL/life= 0.1975 ESAL/axle*78 axles/da*5da/wk*52 wks/yr*38.44GF=153,964. The ESALs add up to 1,021,662over 25 years. See the entries in the table above.
- Flexible pavement design. Carry out a three-layer flexible pavement for CR600E. The County specifications are: Reliability = 90 percent, S.D. = 0.30, PSI = 1.9, MR values of 15,700 psi (base), 12,280 psi (subbase), and 5,250 psi (subgrade). Regardless of your answer to Problem 1, use 1 million ESALs in this problem.
- (12 points) Find SN1, SN2, and SN3 using the design chart in FTE Figure 9.15. Make a copy (enlarged if possible) of the design chart and attach it to your solutions. Remember to label each scale with the value you used.
SN3=3.7, SN2=2.7, SN1=2.5. See design chart on next page.
- (12 points) Using the steps on FTE page 482 and material coefficients a1=0.35, a2=0.19, a3=0.09, what are d1, d2, and d3? Do not violate the minimum thickness values in FTE Table 9.7.
SN1=a1*d1; d1=SN1/a1=2.5/0.35=7.14 in.7.5 in., which is more than the minimum of 3 in. required for HMA and 1.02 million ESALs in Table 9.7.
SN2=a1d1+a2d2=(0.35*7.5)+(0.19*d2)=2.7; d2=0.0.075/0.19=0.395 in.0.5 in., but Table 9.7 requires d2=6 in.
SN3=a1d1+a2d2+a3d3=(0.35*7.5)+(0.19*6.0)+(0.09*d3)=3.7. d3=-0.065/0.09=-0.72 in. This is less than zero, so the major increase in base material (d2) makes the subbase (d3) unnecessary.
- (8 points) Using the material costs in FTE Table 9.8, calculate the cost per lane-mile-inch for each layer that you designed in Part B, then compute the total pavement cost for one lane-mile. Include excavation and earthwork costs at $17.08/CY.
Total pavt thickness = d1+d2+d3=7.5+6.0+0.0=13.5 in. 12 ft wide*(13.5/12) ft deep*5280 ft/mi*cu yd/27 cu ft *$17.08/cu yd = $45,091 to excavate.
7.5 in. Asph Concr * $39,352.50/in. = $295,143.75.
6 in. base * $5791.50/in. = $34,749.00
Total cost per lane-mi. = $45,091+$295,143.75+$34,749.00 = $374,983.75.
(If “Alternate base” in FTE Table 9.8a is used,
6 in. base * $15,480/in. = $95,040 and
Total cost per lane-mi. = $45,091 +$295,143.75+$95,040 = $435,274.75.)
- (8 points) Using the minimum thicknesses of d1 and d2 allowed in FTE Table 9.7, show how to find d3. How much would this pavement design cost to construct? Include excavation and earthwork costs at $17.08/CY.
Reduce d1 from 7.5 in. to 3.0 in.
SN2=a1d1+a2d2=(0.35*3.0)+(0.19*d2)=2.7; d2=1.65/0.19=8.68 in 9.0 in., more than the 6 in. that Table 9.7 requires. Use d2*=6.0 and proceed.
SN3=a1d1+a2d2+a3d3=(0.35*3.0)+(0.19*6.0)+(0.09*d3)=3.7. d3=1.51/0.09=16.77 in. d3*=17.0 in.
The pavement is now 3.0+6.0+17.0= 26.0 in. thick, so the excavation costs increase: 12 ft wide*(26/12) ft deep*5280 ft/mi*cu yd/27 cu ft *$17.08/cu yd = $86,842 to excavate.
3 in. Asph Concr * $39,352.50/in. = $118,057.50.
6.0 in. base * $5791.50/in. = $34,749.00
17.0 in. subbase *$2656.50/in. = $46,160.50
Total cost per lane-mi. = $86,842+$118,057.50+$34,749.00+$46,160.50= $285,809.
(If “Alternate base” in FTE Table 9.8a is used,
6 in. base * $15,480/in. = $95,040 and
Total cost per lane-mi. = $86,842 +$118,057.50+$95,040 +$46,160.50= $346,100.)
Extra excavation, but thinner expensive layers, lowers total cost. Note: If we left d2*=9.0 in., then d3*=10.5 in. and the cost per lane-mi. would be = $75,152 +$117,869+$52,040+$27,849 = $272,909. Cheaper than d2*=6.0 in. and d3*=16.5 in.!
- (20 points) Rigid pavement design. The County asks you to also design a rigid pavement to/from the ethanol plant. Use the same specifications, except as follows: S’c = 828 psi, J = 3.0, Ec = 4.7*106 psi, k = 59 pci, and Cd = 1.0. Regardless of your answer to Problem 1, use 1 million ESALs in this problem. Using the design chart in FTE Figures 9.26 and 9.27, how thick must the concrete slab be? Make a copy (enlarged if possible) of the design chart and attach it to your solutions. Remember to label each scale (and the Match Line) with the value you used.
See design chart below. After the first turning line in the Materials Portion of the Design Chart, the lines drawn for this problem follow closely the lines for the printed example. The vertical and horizontal lines in the design grid cross at a point between the 5” and 6” lines, but closer to D=6” than to D=5”, so the concrete slab design thickness is 6.0 inches.